3
$\begingroup$

How can one evaluate the indefinite integral $$\int \frac{\sqrt{x}+ 1}{\sqrt{x}-1} \,dx$$

First I tried to multiply by $\frac{\sqrt{x}+1}{\sqrt{x}+1}$, but I couldn't get the final result. I also tried to seperate the integral into two $$\int \frac{\sqrt{x}}{\sqrt{x}-1} dx + \int \frac{1}{\sqrt{x}-1} dx$$

which failed to lead to a solution either. Is there another method that can be used?

$\endgroup$
3
  • 2
    $\begingroup$ What about the substitution $\sqrt x=t \Rightarrow x=t^2 \Rightarrow dx=2t\;dt$ ? $\endgroup$
    – User
    Commented Oct 5, 2021 at 21:31
  • 2
    $\begingroup$ @MasterShifu I did it to, thank you. $\endgroup$
    – Student
    Commented Oct 5, 2021 at 21:35
  • 2
    $\begingroup$ You're welcome. By the way if you want to separate the fraction first, I would suggest writing it as $\frac{(\sqrt x-1)+2}{\sqrt x-1}=1+\frac2{\sqrt x-1}$. $\endgroup$
    – User
    Commented Oct 5, 2021 at 21:47

3 Answers 3

Reset to default
4
$\begingroup$

Hint The form of the denominator suggests the substitution $$t = \sqrt{x} - 1 .$$ In particular, $x$ can be expressed as a polynomial in $t$, so the substitution yields a rational integrand with denominator $t$.

$\endgroup$
2
  • $\begingroup$ You're right, it really works, thanks. $\endgroup$
    – Student
    Commented Oct 5, 2021 at 22:03
  • 2
    $\begingroup$ @Student you have to take special care when $0 > t$, since then, $\log(t)$ is undefined. When $0 > t$, $\int \frac{dt}{t} = \int \frac{-dt}{-t} = - \log(-t).$ $\endgroup$
    – user2661923
    Commented Oct 5, 2021 at 22:43
2
$\begingroup$

\begin{align} &\int \frac{\sqrt{x}+ 1}{\sqrt{x}-1} \,dx\\ =& \int \frac{x+\sqrt{x}}{\sqrt x(\sqrt{x}-1)} \,dx =\int \frac{(\sqrt{x}+2)(\sqrt x-1)+2}{\sqrt x(\sqrt{x}-1)} \,dx\\ =& \int 1+\frac2{\sqrt x}+\frac{2}{\sqrt x(\sqrt{x}-1)} \,dx =x +4\sqrt x+4\ln|\sqrt x-1|+C \end{align}

$\endgroup$
1
$\begingroup$

$$ \begin{aligned} \int \frac{\sqrt{x}+1}{\sqrt{x}-1} d x & =2 \int \frac{x+\sqrt{x}}{\sqrt{x}-1} d(\sqrt{x}) \\ & =2 \int \frac{y^2+y}{y-1} d y, \textrm{ where } y=\sqrt{x } \\ & =2 \int\left(y+2+\frac{2}{y-1}\right) d y \\ & =y^2+4 y+4 \ln |y-1|+C \\ & =x+4 \sqrt{x}+4 \ln |\sqrt{x}-1|+C \end{aligned} $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .